Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $k = \dfrac{n^2 - 5n - 50}{n - 1} \div \dfrac{n^2 - 10n}{n - 1} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $k = \dfrac{n^2 - 5n - 50}{n - 1} \times \dfrac{n - 1}{n^2 - 10n} $ First factor the quadratic. $k = \dfrac{(n - 10)(n + 5)}{n - 1} \times \dfrac{n - 1}{n^2 - 10n} $ Then factor out any other terms. $k = \dfrac{(n - 10)(n + 5)}{n - 1} \times \dfrac{n - 1}{n(n - 10)} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac{ (n - 10)(n + 5) \times (n - 1) } { (n - 1) \times n(n - 10) } $ $k = \dfrac{ (n - 10)(n + 5)(n - 1)}{ n(n - 1)(n - 10)} $ Notice that $(n - 1)$ and $(n - 10)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac{ \cancel{(n - 10)}(n + 5)(n - 1)}{ n(n - 1)\cancel{(n - 10)}} $ We are dividing by $n - 10$ , so $n - 10 \neq 0$ Therefore, $n \neq 10$ $k = \dfrac{ \cancel{(n - 10)}(n + 5)\cancel{(n - 1)}}{ n\cancel{(n - 1)}\cancel{(n - 10)}} $ We are dividing by $n - 1$ , so $n - 1 \neq 0$ Therefore, $n \neq 1$ $k = \dfrac{n + 5}{n} ; \space n \neq 10 ; \space n \neq 1 $